When the concentration of 'A' is 0.1 M, it decomposes to give 'X' by a first-order process with a rate constant of 6.93 × 10^-2 min^-1. The reactant 'A', in the presence of catalyst, gives 'Y' by a se

Question

When the concentration of 'A' is 0.1 M, it decomposes to give 'X' by a first-order process with a rate constant of 6.93 × 10^-2 min^-1. The reactant 'A', in the presence of catalyst, gives 'Y' by a second-order mechanism with the rate constant of 0.2 min^-1 M^-1. In order to make half-life of both the processes, same, one should start the second-order reaction with an initial concentration of 'A' equal to

Prashant Kotian · Accepted Answer

Correct answer: D. For the first-order reaction, t_1/2 = 0.693 / (6.93 × 10^-2) = 10 min. For the second-order reaction to have the same t_1/2, t_1/2 = 1 / (K [A0]) = 10. Thus, [A0] = 1 / (0.2 × 10) = 0.5 M. Therefore, correct answer is D.

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